Calculus Cheat Sheet
Derivatives
Definition and Notation
f ( x + h) − f ( x)
If y = f ( x ) then the derivative is defined to be f ′ ( x ) = lim
.
h →0
h
If y = f ( x ) then all of the following are
equivalent notations for the derivative.
df dy d
f ′ ( x ) = y′ =
=
= ( f ( x ) ) = Df ( x )
dx dx dx
If y = f ( x ) then,
If y = f ( x ) all of the following are equivalent
notations for derivative evaluated at x = a .
df
dy
f ′ ( a ) = y ′ x =a =
=
= Df ( a )
dx x =a dx x =a
Interpretation of the Derivative
2. f ′ ( a ) is the instantaneous rate of
1. m = f ′ ( a ) is the slope of the tangent
change of f ( x ) at x = a .
line to y = f ( x ) at x = a and the
3. If f ( x ) is the position of an object at
time x then f ′ ( a ) is the velocity of
equation of the tangent line at x = a is
given by y = f ( a ) + f ′ ( a )( x − a ) .
the object at x = a .
Basic Properties and Formulas
If f ( x ) and g ( x ) are differentiable functions (the derivative exists), c and n are any real numbers,
1.
( c f )′ = c f ′ ( x )
2.
( f ± g )′ = f ′ ( x ) ± g ′ ( x )
3.
( f g )′ =
 f
4. 
g
d
(c) = 0
dx
d n
6.
x ) = n x n−1 – Power Rule
(
dx
d
7.
f ( g ( x )) = f ′ ( g ( x )) g′ ( x )
dx
This is the Chain Rule
5.
f ′ g + f g ′ – Product Rule
′ f ′ g − f g ′
– Quotient Rule
 =
g2

(
)
Common Derivatives
d
( x) = 1
dx
d
( sin x ) = cos x
dx
d
( cos x ) = − sin x
dx
d
( tan x ) = sec2 x
dx
d
( sec x ) = sec x tan x
dx
d
( csc x ) = − csc x cot x
dx
d
( cot x ) = − csc2 x
dx
d
1
sin −1 x ) =
(
dx
1 − x2
d
1
cos −1 x ) = −
(
dx
1 − x2
d
1
tan −1 x ) =
(
dx
1 + x2
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
d x
a ) = a x ln ( a )
(
dx
d x
e ) = ex
(
dx
d
1
ln ( x ) ) = , x > 0
(
dx
x
d
1
( ln x ) = x , x ≠ 0
dx
d
1
log a ( x ) ) =
, x>0
(
dx
x ln a
© 2005 Paul Dawkins
Calculus Cheat Sheet
Chain Rule Variants
The chain rule applied to some specific functions.
n
n −1
d
d
1.
 f ( x )  = n  f ( x )  f ′ ( x )
5.
cos  f ( x )  = − f ′ ( x ) sin  f ( x ) 
dx
dx
d f ( x)
d
e
= f ′ ( x ) e f ( x)
tan  f ( x )  = f ′ ( x ) sec 2  f ( x ) 
2.
6.
dx
dx
d
′
f ( x)
d
7.
( sec [ f ( x)]) = f ′( x) sec [ f ( x)] tan [ f ( x)]
3.
ln  f ( x )  =
dx
dx
f ( x)
f ′( x)
d
d
tan −1  f ( x )  =
8.
2
4.
sin  f ( x )  = f ′ ( x ) cos  f ( x ) 
dx
1 +  f ( x ) 
dx
)
(
(
)
(
(
)
(
)
)
(
(
)
)
Higher Order Derivatives
The Second Derivative is denoted as
The nth Derivative is denoted as
d2 f
dn f
f ′′ ( x ) = f ( 2) ( x ) = 2 and is defined as
f ( n ) ( x ) = n and is defined as
dx
dx
′
f ′′ ( x ) = ( f ′ ( x ) )′ , i.e. the derivative of the
f ( n ) ( x ) = f ( n −1) ( x ) , i.e. the derivative of
first derivative, f ′ ( x ) .
the (n-1)st derivative, f ( n−1) ( x ) .
(
)
Implicit Differentiation
2 x −9 y
3 2
′
Find y if e
+ x y = sin ( y ) + 11x . Remember y = y ( x ) here, so products/quotients of x and y
will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to
differentiate as normal and every time you differentiate a y you tack on a y′ (from the chain rule).
After differentiating solve for y′ .
e 2 x −9 y ( 2 − 9 y′ ) + 3 x 2 y 2 + 2 x3 y y′ = cos ( y ) y′ + 11
2e
2 x −9 y
− 9 y′e
( 2 x y − 9e x
3
2 x −9 y
2 −9 y
+ 3x y + 2 x y y′ = cos ( y ) y′ + 11
2
2
3
− cos ( y ) ) y′ = 11 − 2e2 x −9 y − 3x 2 y 2
⇒
11 − 2e 2 x −9 y − 3x 2 y 2
y′ = 3
2 x y − 9e2 x −9 y − cos ( y )
Increasing/Decreasing – Concave Up/Concave Down
Critical Points
x = c is a critical point of f ( x ) provided either
1. f ′ ( c ) = 0 or 2. f ′ ( c ) doesn’t exist.
Concave Up/Concave Down
1. If f ′′ ( x ) > 0 for all x in an interval I then
f ( x ) is concave up on the interval I.
Increasing/Decreasing
1. If f ′ ( x ) > 0 for all x in an interval I then
2. If f ′′ ( x ) < 0 for all x in an interval I then
2. If f ′ ( x ) < 0 for all x in an interval I then
Inflection Points
x = c is a inflection point of f ( x ) if the
3. If f ′ ( x ) = 0 for all x in an interval I then
concavity changes at x = c .
f ( x ) is increasing on the interval I.
f ( x ) is decreasing on the interval I.
f ( x ) is concave down on the interval I.
f ( x ) is constant on the interval I.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
Calculus Cheat Sheet
Absolute Extrema
1. x = c is an absolute maximum of f ( x )
Extrema
Relative (local) Extrema
1. x = c is a relative (or local) maximum of
f ( x ) if f ( c ) ≥ f ( x ) for all x near c.
2. x = c is an absolute minimum of f ( x )
2. x = c is a relative (or local) minimum of
f ( x ) if f ( c ) ≤ f ( x ) for all x near c.
if f ( c ) ≥ f ( x ) for all x in the domain.
if f ( c ) ≤ f ( x ) for all x in the domain.
Fermat’s Theorem
If f ( x ) has a relative (or local) extrema at
x = c , then x = c is a critical point of f ( x ) .
Extreme Value Theorem
If f ( x ) is continuous on the closed interval
[ a, b] then there exist numbers c and d so that,
1. a ≤ c, d ≤ b , 2. f ( c ) is the abs. max. in
[ a, b] , 3. f ( d ) is the abs. min. in [ a, b] .
Finding Absolute Extrema
To find the absolute extrema of the continuous
function f ( x ) on the interval [ a, b ] use the
following process.
1. Find all critical points of f ( x ) in [ a, b] .
2. Evaluate f ( x ) at all points found in Step 1.
3. Evaluate f ( a ) and f ( b ) .
4. Identify the abs. max. (largest function
value) and the abs. min.(smallest function
value) from the evaluations in Steps 2 & 3.
1st Derivative Test
If x = c is a critical point of f ( x ) then x = c is
1. a rel. max. of f ( x ) if f ′ ( x ) > 0 to the left
of x = c and f ′ ( x ) < 0 to the right of x = c .
2. a rel. min. of f ( x ) if f ′ ( x ) < 0 to the left
of x = c and f ′ ( x ) > 0 to the right of x = c .
3. not a relative extrema of f ( x ) if f ′ ( x ) is
the same sign on both sides of x = c .
2nd Derivative Test
If x = c is a critical point of f ( x ) such that
f ′ ( c ) = 0 then x = c
1. is a relative maximum of f ( x ) if f ′′ ( c ) < 0 .
2. is a relative minimum of f ( x ) if f ′′ ( c ) > 0 .
3. may be a relative maximum, relative
minimum, or neither if f ′′ ( c ) = 0 .
Finding Relative Extrema and/or
Classify Critical Points
1. Find all critical points of f ( x ) .
2. Use the 1st derivative test or the 2nd
derivative test on each critical point.
Mean Value Theorem
If f ( x ) is continuous on the closed interval [ a, b ] and differentiable on the open interval ( a, b )
then there is a number a < c < b such that f ′ ( c ) =
f (b) − f ( a )
.
b−a
Newton’s Method
If xn is the nth guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn +1 = xn −
provided f ′ ( xn ) exists.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
f ( xn )
f ′ ( xn )
© 2005 Paul Dawkins
Calculus Cheat Sheet
Related Rates
Sketch picture and identify known/unknown quantities. Write down equation relating quantities
and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time
you differentiate a function of t). Plug in known quantities and solve for the unknown quantity.
Ex. A 15 foot ladder is resting against a wall.
Ex. Two people are 50 ft apart when one
The bottom is initially 10 ft away and is being
starts walking north. The angle θ changes at
0.01 rad/min. At what rate is the distance
pushed towards the wall at 14 ft/sec. How fast
between them changing when θ = 0.5 rad?
is the top moving after 12 sec?
x′ is negative because x is decreasing. Using
Pythagorean Theorem and differentiating,
x 2 + y 2 = 152 ⇒ 2 x x′ + 2 y y′ = 0
After 12 sec we have x = 10 − 12 ( 14 ) = 7 and
so y = 152 − 7 2 = 176 . Plug in and solve
for y′ .
7
7 ( − 14 ) + 176 y′ = 0 ⇒ y′ =
ft/sec
4 176
We have θ ′ = 0.01 rad/min. and want to find
x′ . We can use various trig fcns but easiest is,
x
x′
sec θ =
⇒ sec θ tan θ θ ′ =
50
50
We know θ = 0.5 so plug in θ ′ and solve.
x′
sec ( 0.5 ) tan ( 0.5 )( 0.01) =
50
x′ = 0.3112 ft/sec
Remember to have calculator in radians!
Optimization
Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for
one of the two variables and plug into first equation. Find critical points of equation in range of
variables and verify that they are min/max as needed.
Ex. We’re enclosing a rectangular field with
Ex. Determine point(s) on y = x 2 + 1 that are
500 ft of fence material and one side of the
closest to (0,2).
field is a building. Determine dimensions that
will maximize the enclosed area.
Minimize f = d 2 = ( x − 0 ) + ( y − 2 ) and the
2
Maximize A = xy subject to constraint of
x + 2 y = 500 . Solve constraint for x and plug
into area.
A = y ( 500 − 2 y )
x = 500 − 2 y ⇒
= 500 y − 2 y 2
Differentiate and find critical point(s).
A′ = 500 − 4 y ⇒ y = 125
nd
By 2 deriv. test this is a rel. max. and so is
the answer we’re after. Finally, find x.
x = 500 − 2 (125 ) = 250
The dimensions are then 250 x 125.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
2
constraint is y = x 2 + 1 . Solve constraint for
x 2 and plug into the function.
2
x2 = y − 1 ⇒ f = x2 + ( y − 2)
= y −1 + ( y − 2) = y 2 − 3 y + 3
Differentiate and find critical point(s).
f ′ = 2y −3
⇒ y = 32
By the 2nd derivative test this is a rel. min. and
so all we need to do is find x value(s).
x 2 = 32 − 1 = 12 ⇒ x = ± 12
2
The 2 points are then
(
1
2
)
(
, 32 and −
1
2
, 32
)
© 2005 Paul Dawkins